DATE POSTED:MONDAY 08/06/2020
CLASS 1:TUESDAY 09/02/2020
Find out which of these are pythagorean triples
(24,58,62)
b) (14,49,50)
C)(15,36,39)
2)Calculate to 2s.f the length of a square that measures 20cm by 20cm
3) A student cycles from home to school first eastwards to a road junction 12km from home,then southwards to school.if the school is 19km from home. How far is it from the road junction,
CLASS 2:DATE WEDNESDAY 10/06/2020
Which of these are pythagoreans triple?
a)(10,24,26)
b)(12,29.32)
c)(14,49,50)
2)A rectangle measures 8cm by 15cm. Use pythagoreans rule to calculate the length of one of its diagonals.
3)A ladder 9.6m long leans against a wall.it touches the wall at a point 9m above the ground .Find the distance of the foot of the ladder from the wall.
DATE POSTED 02/06/2020
CORRECTION OF LAST WEEK ASSIGNMENT
Thursday 28/05/2020
1) Solve
3(2X-1)/4=4(X+2)/3-3
Solution
Multiply LHS and RHS by the lcm of 12
12*3(2X-1)/4=12*4(X+2)/3-12*3
This is simplified to give,
3*3(2X-1)=4*4(X+2)-36
This gives us,
9(2X-1)=16(X+2)-36
Opening the bracket we have,
18X-9=16X+32-36
18X-9=16X-4,
Collecting like terms,
18X-16X=-4+9,
This gives us,
2X=5
Divide both sides by 2,
2X/2=5/2
This gives,
X=2.5
2)Let the number be Y
Subtract 17 from the number gives us,
Y-17
Divide the result by 5 we have,
(Y-17)/5
The final answer is 3,
This gives us,
(Y-17)/5=3,
On cross multiplication,
(Y-17)=3*5
(Y-17)=15
Collecting like terms
Y=15+17
Y=32
3)Let the number be Z
Subtract 14 from the number gives,
Z-14,
With the difference trebled,
3(Z-14),………………………………………(1)
Two third of the number is
(2/3)Z………………………………………………(2)
The result in (1) is equal to (2)
3(Z-14)=(2/3)Z
Multiply both sides by LCM of 3,
3*3(Z-14)=3*(2/3)Z,
This is simplified to give
9(Z-14)=2Z
Opening the bracket we have,
9Z-126=2Z
Collecting like terms,
9Z-2Z=126
7Z=126
Divide both sides by 7,
7Z/7=126/7
This gives ,
Z=18
DATE WEDNESDAY 27/05/2020
WORD PROBLEMS INVOLVING SOLVING EQUATIONS
A number is multiplied by 6 and then 4 is added. The result is 34, find the number.
Solution
Let the number be X.
The number multiplied by X gives
6*X=6X……………………..(1)
Addition of 4 to (1) gives
6X+4…………………………(2)
The result of (2) gives 34
Then we have
6X+4=34…………………..(3)
Subtract 4 from both sides of the equations
6X+4-4=34-4
This gives
6X=30
Divide both equations by 6
6X/6=30/6
This gives
X=5
Example 2
I am thinking of a number. I take away 5.The result is 14.Wnat number did I think of?
Solution
Let the number be Y.
Taking away 5 from the number gives
Y-5……………………….(1)
The result of (1) is 14
Y-5=14
Add 5 to both sides of the equations
Y-5+5=14+5
This gives
Y=19
ASSIGNMENT
- I think of a number .I double it the result is 96. Find the number.
- When I add a number to another number four times as big, the result is 30.Find the number.
- I add 12 to a certain number and then double the result. The answer is 42. Find the original number
- I subtract 8 from a certain number. I then multiply the result by 3. The final answer is 21. Find the original number
DATE: THURSDAY 25/05/2020
WORD PROBLEMS INVOLVING SOLVING EQUATIONS 2
Example1
I think of a number .I double it .I divide the result by 5.My answer is 6 .What number did I think of?
Solution
Let the number be M
Doubling the number gives
2*M=2M…………………….(1)
Dividing (1) by 5 gives
2M/5
The answer is 6
2M/5=6
Cross multiply
2M=6*5
2M=30…………………(2)
Divide by (2)
2M/2=30/2
M=15
Example 2
I add 45 to a certain number and then divide the sum by 2. The result is 5 times the original number. Find the number.
Solution
Let the number be Z
Addition of 45 to Z gives
45+Z……………………………(1)
Divide (1) by 2 gives
(45+Z)/2………………………(2)
Five times the original number gives
5Z
The result (1) IS equal to (2)
(45+Z)/2=5Z
Cross multiply
45+Z=2*5Z
This gives
45+Z=10Z
Subtract Z from both sides of the equations
45+Z-Z=10Z-Z
This gives
45=9Z
Divide both sides by 9
45/9=9Z/9
This gives
Z=5
ASSIGNMENT
Solve these equations
1)b+3/5=3b+3/3
2)3(2x-1)/4=4(x+2)/3-3
3)i subtract 17 from a certain number and then divide the result by 5.My final answer is 3. What was the original number?
4) I think of a number. I subtract 14 from the number and then treble the difference .The result is two-thirds of the number. Find the original number
Correction to last assignment
Wednesday 27/05/2020
Let the number be X
When the number is double that gives 2*X
If the result is 96,
That gives us,
2X=96
Divide both sides by 2,
2X/2=96/2
X=48
2)Let the number be Z
Four times the number=4*X
Addition of the number with 4 times the number equals
X +4X
The result is 20
This gives us
X+4X=20 ,
5X=20
Divide both sides by 5
5X/5=20/5
X=4
3)Let the number be Y
Addition of 12 to the number gives
Y+12
Doubling the result gives
(Y+12)2
The answer is 42
2(Y+12)=42
Opening the bracket
2Y+24=42
Collect like terms,
2Y=42-24
2Y=18
Divide both sides by 2,
2Y/2=18/2
Y=9
4)Let the number be M
Subtract 8 from the number gives
M-8
Multiply the result by 3
(M-8)*3
The final answer is 21
This gives us
3(M-8)=21
Opening brackets.
3M-24=21
Collect like terms
3M=21+24
This gives
3M=45
Divide both sides by 3
3M/3=45/3
M=15